Maclaurin series: RI 2024 Promo Q6

(a)[5] $$\mathrm{e}^y=1+\sin3x\quad------(1)$$ Differentiating w.r.t. $x$, $$\mathrm{e}^y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}=3\cos3x.$$ Differentiating w.r.t. $x$, $$\begin{aligned} \mathrm{e}^y\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}+\mathrm{e}^y\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\frac{\,\mathrm{d} y}{\,\mathrm{d} x} &=-9\sin3x=-9(\mathrm{e}^y-1)\quad\text{(from (1))},\\ \mathrm{e}^y\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}+\mathrm{e}^y\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^2+9\mathrm{e}^y&=9. \end{aligned}$$ Dividing throughout by $\mathrm{e}^y$, $$\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}+\left(\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right)^2+9=9\mathrm{e}^{-y}\quad\text{(shown)}.$$ Differentiating w.r.t. $x$, $$\frac{\,\mathrm{d}^3y}{\,\mathrm{d} x^3}+2\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2} =9\mathrm{e}^{-y}\left(-\frac{\,\mathrm{d} y}{\,\mathrm{d} x}\right).$$ When $x=0$, $$y=0;\quad\frac{\,\mathrm{d} y}{\,\mathrm{d} x}=3;\quad\frac{\,\mathrm{d}^2y}{\,\mathrm{d} x^2}=-9;\quad\frac{\,\mathrm{d}^3y}{\,\mathrm{d} x^3}=27.$$ $$\Rightarrow y=0+3x+\frac{-9}{2!}x^2+\frac{27}{3!}x^3+\cdots$$ $$\therefore y=3x-\frac92x^2+\frac92x^3+\cdots$$

(b)[3] $$\begin{aligned} \mathrm{e}^y&=1+\sin3x\\ \Rightarrow y&=\ln(1+\sin3x)\\ &=\ln\left(1+\left(3x-\frac{(3x)^3}{3!}\right)+\cdots\right)\\ &=\left(3x-\frac{(3x)^3}{3!}\right)-\frac{\left(3x-\frac{(3x)^3}{3!}\right)^2}{2} +\frac{\left(3x-\frac{(3x)^3}{3!}\right)^3}{3}+\cdots\\ &=3x-\frac{27x^3}{6}-\frac{9x^2}{2}+\frac{27x^3}{3}+\cdots\\ &=3x-\frac92x^2+\frac92x^3+\cdots \end{aligned}$$ which is same as the expansion for $y$ found in part (a), up to and including the term in $x^3$.